Chapter 4.4
1 Since the wavelength is of the same order of magnitude as the opening there will be
appreciable diffraction as in Fig.4.2 of the textbook on page 239.
2 The wavelength is now very small compared to the aperture size and so no
appreciable diffraction will take place – see Fig. 4.1 on page 238 of the textbook.
3 Reflection and diffraction. By placing a mirror somewhere.
4 At M the waves from the two speakers arrive without any path difference and so
interfere constructively leading to a loud sound. At P there is a path difference
between the two waves equal to 4.00 m. When this path difference is a half integral of
1
the wavelength we will get destructive interference. Thus 4.00 = (n + )! . The
2
largest wavelength corresponds to n = 0 and so ! = 8.00 m .
5 There is poor reception because of destructive interference between the waves
reaching the antenna directly and those reflecting off the mountain. Assuming no
phase shifts upon reflection, the path difference is double the distance between the
1
house and the mountain. Then 2d = (n + )! . The smallest d corresponds to n = 0
2
and so d = 400 m .
3.0 " 10 8
= 3.3 m . The path difference at
6 The wavelength of the radio waves is ! =
90 " 10 6
1
A is zero and at B it is 2(AB) . Then 2(AB) = (n + )! and so the minimum distance
2
!
AB is AB = = 0.83 m .
4
340
= 0.40 m . (a) The path difference is
850
path difference 0.80
9.0 ! 8.2 = 0.80 m .
=
= 2 , an integer, so we have constructive
!
0.40
interference, a loud sound. (b) The path difference now is 8.7 ! 8.1 = 0.60 m and so
path difference 0.60
=
= 1.5 , a half integer and so destructive interference, no sound.
!
0.40
7 The wavelength of sound is ! =
8 Superposition means that when two waves arrive at the same point in space the
resulting displacement of the wave is the algebraic sum of the displacements of the
individual waves. For illustrations see diagrams in textbook on page 241.